/*
Problem Description
Huatuo was a famous doctor. He use identical bottles to carry the medicine. There are different types of medicine. Huatuo put medicines into the bottles and chain these bottles together.
However, there was a critical problem. When Huatuo arrived the patient's home, he took the chain out of his bag, and he could not recognize which bottle contains which type of medicine, but he remembers the order of the bottles on the chain.
Huatuo has his own solution to resolve this problem. When he need to bring 2 types of medicines, E.g. A and B, he will put A into one bottle and put B into two bottles. Then he will chain the bottles in the order of ′−B−A−B−′. In this way, when he arrived the patient's home, he knew that the bottle in the middle is medicine A and the bottle on two sides are medicine B.
Now you need to help Huatuo to work out what's the minimal number of bottles needed if he want to bring N types of medicine.
Input
The first line of the input gives the number of test cases, T(1≤T≤100). T lines follow. Each line consist of one integer N(1≤N≤100), the number of types of the medicine.
Output
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimal number of bottles Huatuo needed.

Sample Input
1
2
Sample Output
Case #1: 3
题目大意：华佗给人治病，拿个很多药瓶，每一个都有药，但是只看瓶子并不能看出是哪种药物，但是有一个链子，可以把所有的药都串成一条线，如果只拿了两种药的时候，
一种药的瓶子是A,另一种是B,此时可以排序为BAB，此时只需要三个瓶子即可，如果是AB的话，并不能区分两瓶都是什么药。查找最小需要多少个药瓶，可以很容易的区分每一种药。
思路：如果有一瓶的时候并不需要区分，只需要一瓶即可，每次增加的时候，都需要两个瓶子分布在这个瓶子的两边，每次都采用这种思路，可保证瓶子的数量最少。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#define PI acos(-1)
#define M(n, m) memset(n, m, sizeof(n));
const int INF = 1e9 + 7;
const int maxn = 1e5 + 100;
using namespace std;

int main()
{
    int num, n;
    scanf("%d", &num);
    for (int k = 1;k <= num;k ++)
    {
        scanf("%d", &n);
        int sum = n * 2 - 1;
        printf("Case #%d: %d\n", k, sum);
    }
    return 0;
}
